Jemurray3 3 years ago Derivative of x^2:

1. Jemurray3

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2. Jemurray3

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3. ash2326

y=x^2 dy/dx=2x we know derivative of x^n=nx^(n-1)

4. Jemurray3

I'm not asking :) hold on

5. Jemurray3

That tangent line has an equation $y - x_0^2 = m(x-x_0)$ for some m. Let x > x0. Then, since the points on the line are less than the points on the curve, $x^2 \geq m(x-x_0) + x_0^2$ $x^2-x_0^2 = (x+x_0)(x-x_0) \geq m(x-x_0)$ and x-x0 is positive, so $x+x_0 \geq m$ Next, for x < x0, the points on the line are likewise less than the points of the curve, so we get the same equation but now x-x0 < 0 so we obtain $x+x_0 \leq m$ now, for the first one, since x must be greater than or equal to x0, we can say that $x+x_0 \geq 2x_0 \geq m$ and Likewise for the second, since x must be less than x0, $x + x_0 \leq 2x_0 \leq m$ thus $2x_0 \leq m \leq 2x_0$ and thus $m = 2x_0$ so the slope of the tangent line to the curve y=x^2 at some point x0 is equal to 2x0, no difference quotients or limits required. Ta da!

6. Jemurray3

Don't judge me, I was bored and decided to try a derivative without a limit. Shush. :)

7. Mahnoor

It's really interesting, actually.